EE130/230M Review Session

1. Small Signal Models for MOSFET/BJT

2. MOS Electrostatics

BJT Small Signal Model

• A small change in IB or VBE will result in a small change in IC and VCE

• input current = be( 1 / )j C r v

output current = m beg v

the cutoff frequency (wT = 2pfT) is defined at

• input voltage = bev

output voltage =

= d1 / dg i

be/m dg g vvbe

r gmvbe

C

E

B

E

C

+

,

,

( )1 / /

1 1

1 / /

c m m

b m dc F m J BE

dc F J BE C

i g g

i r j C g j g C

j C kT qI

Example

A BJT is biased at IC = 1 mA and VCE = 3V. bdc = 90, tF = 5ps, T = 300K. Find (a) gm , (b) rp , (c) Cp .

Solution:

(a)

(b) rp = bdc / gm = 90/0.039 = 2.3 kW

(c)

siemens)(milliqkTIg Cm mS 39V

mA39

mV 26

mA 1)//(

ad)(femto fargC mF fF 19F109.1039.0105 1412

EE130/230M Spring 2013 Lecture 27, Slide 3

MOSFET Small Signal Model

• A small change in VG will result in a small change in ID and VDS

• input current = g gj C v

output current = m gg v

m g

T g g

12

g v

f C v

the cutoff frequency (wT = 2pfT) is defined at

• input voltage = gv

output voltage =

= d1 / dg i

g/m dg g v

Summary of Small Signal Models

• Inputs/Outputs

• Linearized equivalent circuits

• Key components

• Performances Gain Cut-off frequency

MOS Capacitor Energy Band Diagram

Since the substrate is n-type, this is a PMOS capacitor, i.e. a sufficiently large negative gate voltage is required to deplete the semiconductor surface of mobile electrons and form an inversion-layer of holes there.

a) Assuming that the n-type Si substrate is non-degenerately doped, its work function is

S = χ + (EC-EF) = χ + EG/2 – (kT/q)∙ln(ND/ni) = 4.05 eV + 0.56 eV – 0.48 eV = 4.13 eV

The gate material is heavily doped p-type silicon with work function χ + EG 5.2 eV

Ideally, the flat-band voltage is given by the difference between the gate and the semiconductor work functions:

qVFB= M – S 5.2 eV – 4.13 eV = 1.07 eV

Therefore the flat-band voltage VFB 1.07 V

Consider an ideal MOS capacitor maintained at T = 300K with the following parameters: • Gate material is p+ polycrystalline-silicon (work function M = 5.2 eV) • Substrate is n-type Si, with doping concentration 1018 cm-3 (assume that this is non-

degenerate) • Oxide thickness xo = 2 nm

a) What is the flat-band voltage, VFB, of this capacitor? b) Sketch the energy-band diagrams, labeling qVG, qS, qVox (no numerical values required), for the

following bias conditions: i) flat band ii) accumulation iii) equilibrium iv) strong inversion

Problem:

Solution:

Flat-band Conditiona) The band diagrams corresponding to the various regions of operation are drawn below:

i) Flat-band condition: VG = VFB so that there is no band-bending

Equilibrium Conditioni) Equilibrium: Metal and Si Fermi levels are equal, VG = 0 V

Note that the surface is depleted of mobile electrons, i.e. there is a depletion region (of width W) at the surface of the Si.

Accumulation Conditioni) Accumulation: VG > VFB so that there are majority carrier electrons accumulated at the Si surface

Strong Inversion Conditioni) Strong inversion: VG is more negative than the threshold voltage VT, so that the surface is strongly p-type,

i.e. with a (mobile) hole density greater than the (immobile) density of ionized donor atoms. The depletion region has a width WT, and the total voltage dropped across the Si F = 2F = -2(kT/q)∙ln(ND/ni)=-0.96 V.

Charge distribution

Charge amount

Capacitance

Equivalent circuits

MOS Capacitor Charge vs. Gate Voltage

Accumulation Depletion Inversion

VT

Flat-band voltageMaximum depletion charge

C (pF)

Vg (V)

70

0.25 -0.7

MOS Capacitance vs. Gate Voltage

Accumulation

DepletionInversion

VTFlat-band voltage

Minimum capacitance